Question

我已经尝试过这样的其他问题，但似乎都没有。如果用户在控制台中对数字进行排序后输入Y，我希望它重复。这是代码：

package compsorter;
import java.util.Scanner;
public class Ascending_Order 
{
    public static void main(String[] args) 
    {
    int n, temp;
    Scanner s = new Scanner(System.in);
    System.out.print("Enter no. of elements you want in array:");
    n = s.nextInt();
    int a[] = new int[n];
    System.out.println("Enter all the elements:");

    for (int i = 0; i < n; i++) 
    {
        a[i] = s.nextInt();
    }
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++) 
        {
            if (a[i] > a[j]) 
            {
                temp = a[i];
                a[i] = a[j];
                a[j] = temp;
            }
        }
    }
    System.out.print("In Ascending Order:");
    for (int i = 0; i < n - 1; i++) 
    {
        System.out.print(a[i] + ",");
    }
    System.out.print(a[n - 1]);
    s.close();
}
}

谢谢！

Answer 1

像这样，

do{
   //code here
   //take input from user 'Y' or 'N'
  }while(condition);

Answer 2

你可以放 do {之后 Scanner s = new Scanner(System.in);

和

} while(s.readLine().equals("Y")) 之前 s.close();

Answer 3

尝试这样做的原因。例如：link

Answer 4

您可以使用while循环执行此操作。以下是您应该拥有的主要方法程序。

public static void main(String[] args) {
  String flag = "Y";
  Scanner s = new Scanner(System.in);
  while (true) {

  if ("Y".equalsIgnoreCase(flag)) {
    int n, temp;

    System.out.print("Enter no. of elements you want in array:");
    n = s.nextInt();
    int a[] = new int[n];
    System.out.println("Enter all the elements:");

    for (int i = 0; i < n; i++) {
      a[i] = s.nextInt();
    }
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (a[i] > a[j]) {
          temp = a[i];
          a[i] = a[j];
          a[j] = temp;
        }
      }
    }
    System.out.print("In Ascending Order:");
    for (int i = 0; i < n - 1; i++) {
      System.out.print(a[i] + ",");
    }
    System.out.print(a[n - 1]);
    System.out.println();
    System.out.print("Do you want to continue Y/N?");
    flag = s.next();

  } else {
    break;
  }
}
s.close();
}

如果用户说Y，请尝试重复此程序

4 个答案: